You wish to test the following claim (HaHa) at a significance
level of α=0.01α=0.01. For the context of this problem,
μd=μ2−μ1μd=μ2-μ1 where the first data set represents a pre-test and
the second data set represents a post-test.
Ho:μd=0
Ha:μd<0
You believe the population of difference scores is normally
distributed, but you do not know the standard deviation. You obtain
pre-test and post-test samples for n=48n=48 subjects. The average
difference (post - pre) is ¯d=−20.2d¯=-20.2 with a standard
deviation of the differences of sd=48.2sd=48.2.
What is the test statistic for this sample? (Report answer accurate
to three decimal places.)
test statistic =
What is the p-value for this sample? (Report answer accurate to
four decimal places.)
p-value =
The statistical software output for this problem is:
Hence,
Test statistic = -2.904
p-value = 0.0028
To calculate the test statistic and p-value for this sample, we will use the one-sample t-test formula:
Test Statistic (t) = (¯d - μd) / (sd / √n)
where: ¯d = sample mean of the differences (post - pre) μd = hypothesized population mean difference (in this case, 0) sd = standard deviation of the differences n = sample size
Given the following values: ¯d = -20.2 μd = 0 sd = 48.2 n = 48
Let's calculate the test statistic first:
t = (-20.2 - 0) / (48.2 / √48) t = -20.2 / (48.2 / 6.9282) t = -20.2 / 6.9502 t ≈ -2.908
Now, to find the p-value, we need to use the t-distribution table or a statistical software. Since the p-value corresponds to the area under the t-distribution curve to the left of the test statistic, we are interested in finding P(t < -2.908).
Using a t-distribution table or a statistical software, we find that the p-value for t < -2.908 is approximately 0.0024 (rounded to four decimal places).
So, the test statistic for this sample is approximately -2.908, and the p-value is approximately 0.0024.
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